TITLE: Why doesn't this example of basis change work?
QUESTION [7 upvotes]: I'm learning about the change of basis in linear algebra, and trying to come up with an example to understand it. But somehow my example below doesn't make sense.
Let $B_1 = ((1,0),(0,1))$ be the standard basis in $R^2$, and $v = (2,3)$ be a vector.
Let $B_2$ be another basis of $R^2$, $B_2 = ((1,-1),(1,1))$. My understanding is that, in this system, the same $v$ above has the coordinate $v = (-1/2, 5/2)$.
Let $T: R^2 \rightarrow R^2$, $T(x,y) = (x+2y, 3x - 2y)$.
With $B_1$, 
$M(T) = 
\begin{bmatrix}
    1 & 2 \\
    3 & -2\\
    \end{bmatrix}$, 
and 
$T(2,3) = 
M(T).v = 
\begin{bmatrix}
    1 & 2 \\
    3 & -2\\
    \end{bmatrix}
.\begin{bmatrix}
    2 \\
    3\\
    \end{bmatrix} 
=\begin{bmatrix}
    8\\
    0\\
    \end{bmatrix}$
With $B_2$, $M(T) = \begin{bmatrix}
    -1 & 3 \\
    5 & 1\\
    \end{bmatrix}$, 
but $T(-1/2,5/2) = (9/2,-13/2)$, and 
$M(T).v = 
\begin{bmatrix}
    -1 & 3 \\
    5 & 1\\
    \end{bmatrix}
.\begin{bmatrix}
    -1/2 \\
    5/2\\
    \end{bmatrix}
=\begin{bmatrix}
    8\\
    0\\
    \end{bmatrix}$
I don't understand why in the calculations using $B_2$, $T(-1/2,5/2)$ is not equal to $M(T).v$, and why $M(T).v=\begin{bmatrix}
    8\\
    0\\
    \end{bmatrix}$. I was expecting that $M(T).v$ would give me something different, because we're using another basis. This "something" could then be "converted" back to $\begin{bmatrix}
    8\\
    0\\
    \end{bmatrix}$ in the standard basis.

REPLY [3 votes]: What you want is to figure out how $A$ will look like in the new basis. The short answer is that it will be $B^{-1}AB$, where $B=\begin{bmatrix}
    1 & 1 \\
    -1 & 1\\
    \end{bmatrix}$.
Explanation:
Let $v=(2,3)^T$ and $x=(-1/2,5/2)^T$.
The vector $x$ of the new basis is the same as $v$ in the old basis, so $v=Bx$. Thus $x=B^{-1}v$.
That is: for converting from the new basis $B_2$ to the standard basis $B_1$,you multiply by $B$, and for converting from standard to new basis, you must multiply by $B^{-1}$.
Now, what happens to $A$ in the new basis?
Let $u$ be a vector in the standard basis, so that $w=B^{-1}u$ is the corresponding vector in the new basis.
Which matrix $X$ should be applied to $w$ that corresponds to multiplying $u$ by $A$?
To figure this out, let's do the following:

Move $w$ to the standard basis by applying $B$. [We get $u$]
Now apply $A$; [We get $Au$]
Now apply $B^{-1}$. [We get the vector in the new basis corresponding to $Au$.]

Thus, the matrix $X$ is $B^{-1}AB$.
Instead, what you've done is to use the matrix $AB$, and then multiplied it with $B^{-1}v$ (the vector in the new basis) and this naturally gives $ABB^{-1}v=Av$, the vector in the standard basis (rather than in the new basis).